1961 AHSME Problems/Problem 8

Problem

Let the two base angles of a triangle be $A$ and $B$ , with $B$ larger than $A$ . The altitude to the base divides the vertex angle $C$ into two parts, $C_1$ and $C_2$ , with $C_2$ adjacent to side $a$ . Then:

$\textbf{(A)}\ C_1+C_2=A+B \qquad \textbf{(B)}\ C_1-C_2=B-A \qquad\\ \textbf{(C)}\ C_1-C_2=A-B \qquad \textbf{(D)}\ C_1+C_2=B-A\qquad \textbf{(E)}\ C_1-C_2=A+B$

Solution

$[asy] draw((0,0)--(120,0)--(40,50)--(0,0)); draw((40,50)--(40,0)); draw((40,4)--(44,4)--(44,0)); label("$B$",(10,5)); label("$A$",(100,5)); label("$C_1$",(47,38)); label("$C_2$",(34,33)); [/asy]$

Noting that side $a$ is opposite of angle $A$ , label the diagram as shown.

From the two right triangles, $B + C_2 = 90$ $A + C_1 = 90$ Substitute to get $B + C_2 = A + C_1$ $B - A = C_1 - C_2$ The answer is $\boxed{\textbf{(B)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1961 AHSME Problems/Problem 8

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